Codeforces_GYM_100741 A
http://codeforces.com/gym/100741/problem/A
A. Queries
0.25 seconds
64 megabytes
standard input
standard output
Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my friend, a mathematician, thinks that it is very fun to play with a sequence of integer numbers. He writes the sequence in a row. If he wants he increases one number of the sequence, sometimes it is more interesting to decrease it (do you know why?..) And he likes to add the numbers in the interval [l;r]. But showing that he is really cool he adds only numbers which are equal some mod (modulo m).
Guess what he asked me, when he knew that I am a programmer? Yep, indeed, he asked me to write a program which could process these queries (n is the length of the sequence):
- + p r It increases the number with index p by r. (
,
)
You have to output the number after the increase.
- - p r It decreases the number with index p by r. (
,
) You must not decrease the number if it would become negative.
You have to output the number after the decrease.
- s l r mod You have to output the sum of numbers in the interval
which are equal mod (modulo m). (
) (
)
The first line of each test case contains the number of elements of the sequence n and the number m. (1 ≤ n ≤ 10000) (1 ≤ m ≤ 10)
The second line contains n initial numbers of the sequence. (0 ≤ number ≤ 1000000000)
The third line of each test case contains the number of queries q (1 ≤ q ≤ 10000).
The following q lines contains the queries (one query per line).
Output q lines - the answers to the queries.
3 4
1 2 3
3
s 1 3 2
+ 2 1
- 1 2
2
3
1
m个 树状数组记录a[i]%m,得值。
#define LL long long
#include <iostream> using namespace std; const int MAXN();
const int N(+);
LL n,m,a[N],q,u,v,w; struct Tree
{
LL mm;
LL val[N];
#define lowbit(x) (x&(-x))
void Update(int now,int x)
{
for(;now<=mm;now+=lowbit(now)) val[now]+=x;
}
LL Query(int x)
{
LL ret=;
for(;x;x-=lowbit(x)) ret+=val[x];
return ret;
}
}tree[]; int main()
{
cin>>n>>m;
for(LL i=;i<m;i++) tree[i].mm=n;
for(LL i=;i<=n;i++)
{
cin>>a[i];
tree[a[i]%m].Update(i,a[i]);
}
cin>>q;
for(char ch[];q--;)
{
cin>>ch>>u>>v;
if(ch[]=='+')
{
tree[a[u]%m].Update(u,-a[u]);
a[u]+=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
} else
if(ch[]=='-')
{
if(a[u]<v) cout<<a[u]<<endl;
else
{
tree[a[u]%m].Update(u,-a[u]);
a[u]-=v;
tree[a[u]%m].Update(u,a[u]);
cout<<a[u]<<endl;
}
} else
if(ch[]=='s')
{
cin>>w;
cout<<tree[w].Query(v)-tree[w].Query(u-)<<endl;
}
}
return ;
}
Codeforces_GYM_100741 A的更多相关文章
随机推荐
- Codeforces Round #206 (Div. 2) 部分题解
传送门:http://codeforces.com/contest/355 A:水题,特判0 int k,d; int main(){ //FIN; while(cin>>k>> ...
- 利用反射实现Servlet公共类的抽取
一次请求的执行过程: 请求:发送请求地址-->到达web.xml中,找到地址对应的servlet类-->通过反射调用该类的构造函数,创建该servlet类的对象-->通过当前对象调用 ...
- Thinkphp5图片上传正常,音频和视频上传失败的原因及解决
Thinkphp5图片上传正常,音频和视频上传失败的原因及解决 一.总结 一句话总结:php中默认限制了上传文件的大小为2M,查找错误的时候百度,且根据错误提示来查找错误. 我的实际问题是: 我的表单 ...
- less---查看文件
- .Net 程序在自定义位置查找托管/非托管 dll 的几种方法
原文:.Net 程序在自定义位置查找托管/非托管 dll 的几种方法 一.自定义托管 dll 程序集的查找位置 目前(.Net4.7)能用的有2种: #define DEFAULT_IMPLEMENT ...
- 洛谷——P2093 零件分组
https://www.luogu.org/problem/show?pid=2093 题目描述 某工厂生产一批棍状零件,每个零件都有一定的长度(Li)和重量(Wi).现在为了加工需要,要将它们分成若 ...
- hdu 1588 Gauss Fibonacci(矩阵嵌矩阵)
题目大意: 求出斐波那契中的 第 k*i+b 项的和. 思路分析: 定义斐波那契数列的矩阵 f(n)为斐波那契第n项 F(n) = f(n+1) f(n) 那么能够知道矩阵 A = 1 1 1 0 ...
- php excel文件导出之phpExcel扩展库
php Excel 文件导出 phpExcel 官网 http://phpexcel.codeplex.com/ /** * 导出特定文件 * 依据详细情况而定 */ public function ...
- android图像处理(3)底片效果
这篇将讲到图片特效处理的底片效果.跟前面一样是对像素点进行处理,算法是通用的. 算法原理:将当前像素点的RGB值分别与255之差后的值作为当前点的RGB值. 例: ABC 求B点的底片效果: B.r ...
- R语言-上海二手房数据分析
案例:通过分析上海的二手房的数据,分析出性价比(地段,价格,未来的升值空间)来判断哪个区位的二手房性价比最高 1.载入包 library(ggplot2) library(Hmisc) library ...